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VashaNatasha [74]
3 years ago
11

How many minutes are in 33.6 days

Chemistry
2 answers:
monitta3 years ago
6 0

1






















































































1


11













qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq


Romashka-Z-Leto [24]3 years ago
6 0

The answer is 48384.

24*60=1440

1440*33.6=48384.  I hope this helps!


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Calculate the number of grams of solute in 814.2mL of 0.227 M calcium acetate
kiruha [24]

Answer:

Mass = 29.23 g

Explanation:

Given data:

Volume of solution = 814.2 mL 814.2/1000 = 0.8142 L)

Molarity of solution = 0.227 M

Mass of solute in gram = ?

Solution:

Molarity = number of moles / volume in L

By putting values,

0.227 M = number of moles / 0.8142 L

Number of moles = 0.227 M × 0.8142 L

Number of moles = 0.184 mol

Mass in gram:

Mass = number of moles × molar mass

Molar mass of calcium acetate = 158.17 g/mol

Mass = 0.184 mol × 158.17 g/mol

Mass = 29.23 g

6 0
3 years ago
Which formula represents the compound commonly known as phosphine
Arlecino [84]
I looked up what is the molecular formula for Phosphine and got this: PH3
Hope this helps! Let my know if this was correct.
8 0
3 years ago
Select the most likely product for this reaction:
Rama09 [41]

Answer:

A

Explanation:

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6 0
2 years ago
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All molecules have the same kinetic energy and hence the same speed.<br><br> O True<br><br> O False
topjm [15]
False...their speed differs
4 0
2 years ago
A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0
3 years ago
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